#1

a) Will I get the same gradient if I ll convert the mm to m and then find ln of each?

b) Lets say (just imagine) we re plotting ln(x/mm) against y/mm, then will I have to convert the mm to m when finding out the gradient? And do i ve to convert to the SI unit (in general) when finding out gradients?

c) The equation is y=kx^n
so if they ll ask for the value of k then:
ln(y) = n ln(x) + ln (k)
then y-intercept is equal to ln (k) and to find k, just anti-log. Is this answer enough?

d) I just want to know (just like maybe i ll become quicker at finding scales) ... how do you look for it when finding the scale like the one in the image? What i do is just look at the first two figures like in ln(x) column to make it easier for me, it works actually coz the third firgure will appear in between the separation.
Just wanna make sure.
0
#2

Just give me an example of one of the rows. How will you go through it. like for the first one, 1/60.0 is 0.0166666...
so will you multiply by 10^2 and just take to 3 sf ? so it comes out to be 1.67 and i ll display the label and units as: 1/v (1/10^2 cm^3) right?
Now tell me exactly how i ll find the gradient for this. I ll have to multiply the two p value by 10^3 right?
and then multiply the two x values by 1/10^2 and then convert it to m^3 by dividing by 100^3 ?
0
4 years ago
#3
(Original post by Daniel Atieh)

a) Will I get the same gradient if I ll convert the mm to m and then find ln of each?

b) Lets say (just imagine) we re plotting ln(x/mm) against y/mm, then will I have to convert the mm to m when finding out the gradient? And do i ve to convert to the SI unit (in general) when finding out gradients?

c) The equation is y=kx^n
so if they ll ask for the value of k then:
ln(y) = n ln(x) + ln (k)
then y-intercept is equal to ln (k) and to find k, just anti-log. Is this answer enough?

d) I just want to know (just like maybe i ll become quicker at finding scales) ... how do you look for it when finding the scale like the one in the image? What i do is just look at the first two figures like in ln(x) column to make it easier for me, it works actually coz the third firgure will appear in between the separation.
Just wanna make sure.
a) yes
b) Impossible to answer. it depends what you have been asked to do. What the question is. etc
c) Yes the intercept is ln k and k is the anti-log. What do you mean "answer enough? It depends what the question was.
d) I'm not sure what you are asking. You usually try to fill the graph paper and choose a scale to do this. It depends on the question.
4 years ago
#4
(Original post by Daniel Atieh)

Just give me an example of one of the rows. How will you go through it. like for the first one, 1/60.0 is 0.0166666...
so will you multiply by 10^2 and just take to 3 sf ? so it comes out to be 1.67 and i ll display the label and units as: 1/v (1/10^2 cm^3) right?
Now tell me exactly how i ll find the gradient for this. I ll have to multiply the two p value by 10^3 right?
and then multiply the two x values by 1/10^2 and then convert it to m^3 by dividing by 100^3 ?
This is not the complete question.
#5
(Original post by Stonebridge)
This is not the complete question.
Please accept my apologies, dammit, i attached the wrong one.
0
4 years ago
#6
(Original post by Daniel Atieh)
Please accept my apologies, dammit, i attached the wrong one.
Hello Daniel.

Can you please post the entire question. Giving you a correct response is very difficult without it.

Thanks.
0
#7
(Original post by uberteknik)
Hello Daniel.

Can you please post the entire question. Giving you a correct response is very difficult without it.

Thanks.
Here is it

Read my earlier post for the question. we need to find 1/v for each. Check my method described earlier to see if i m correct please.

Thank you.
0
4 years ago
#8
(Original post by Daniel Atieh)
Here is it

Read my earlier post for the question. we need to find 1/v for each. Check my method described earlier to see if i m correct please.

Thank you.
You have over-complicated things somewhat.

The question is about Boyles law which says that provided the temperature is kept constant then:

Pressure x Volume = constant

pV = k

The experiment produced results for pressure and volume.

Plotting pressure against 1/volume produces a straight line because:

Gradient = rise in y / distance x

so substituting experiment parameters (pressure and 1/v for the x and y axis):

Gradient = pressure / (1 / volume) or

k = p/ (1/v)

i.e. tidying up

k = pV

Since pV = Nmetres

{dimensionally:
pressure = force/area = MLs -2/L2
ML-1s-2
pressure x volume = ML-1s-2 x L3

pV = ML2s-2 in other words MLs-1 x L

i.e. force x distance}

then

k has the dimensions of energy (work) in Joules as long as the units are consistent. i.e. L

L = metres
M = Kg
s = seconds.

You should now see where the final part of the question is headed.
1
#9
(Original post by uberteknik)
You have over-complicated things somewhat.

The question is about Boyles law which says that provided the temperature is kept constant then:

Pressure x Volume = constant

pV = k

The experiment produced results for pressure and volume.

Plotting pressure against 1/volume produces a straight line because:

Gradient = rise in y / distance x

so substituting experiment parameters (pressure and 1/v for the x and y axis):

Gradient = pressure / (1 / volume) or

k = p/ (1/v)

i.e. tidying up

k = pV

Since pV = Nmetres

{dimensionally:
pressure = force/area = MLs -2/L2
ML-1s-2
pressure x volume = ML-1s-2 x L3

pV = ML2s-2 in other words MLs-1 x L

i.e. force x distance}

then

k has the dimensions of energy (work) in Joules as long as the units are consistent. i.e. L

L = metres
M = Kg
s = seconds.

You should now see where the final part of the question is headed.
Thank you so much seriously!
0
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